Question

Use the module from Exercise \(4.24\) to make a program for solving the problems below. 1\. What is the probability of getting two heads when flipping a coin five times? This probability corresponds to \(n=5\) events, where the success of an event means getting head, which has probability \(p=1 / 2\), and we look for \(x=2\) successes. 2\. What is the probability of getting four ones in a row when throwing a die? This probability corresponds to \(n=4\) events, success is getting one and has probability \(p=1 / 6\), and we look for \(x=4\) successful events. 3\. Suppose cross country skiers typically experience one ski break in one out of 120 competitions. Hence, the probability of breaking a ski can be set to \(p=1 / 120\). What is the probability \(b\) that a skier will experience a ski break during five competitions in a world championship? This question is a bit more demanding than the other two. We are looking for the probability of \(1,2,3,4\) or 5 ski breaks, so it is simpler to ask for the probability \(c\) of not breaking a ski, and then compute \(b=1-c\). Define "success" as breaking a ski. We then look for \(x=0\) successes out of \(n=5\) trials, with \(p=1 / 120\) for each trial. Compute \(b .\) Name of program file: binomial_problems.py.

Short answer

1. Probability: \(\frac{5}{16}\). 2. Probability: \(\frac{1}{1296}\). 3. Probability: 0.041.

Step-by-step solution

Understanding Question 1

This involves calculating the probability of getting exactly 2 heads when flipping a coin 5 times. This scenario represents a binomial distribution problem where the number of trials, \(n = 5\), the probability of success (getting heads) in each trial \(p = \frac{1}{2}\), and we are looking for \(x = 2\) successes (heads).

Calculating Probability for Question 1

For a binomial distribution, the probability \(P\) of getting exactly \(x\) successes in \(n\) trials is given by: \[ P(X = x) = \binom{n}{x}p^x(1-p)^{n-x} \] For 2 heads, \(\binom{5}{2} = 10\), so \(P(X = 2) = 10 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^3 = \frac{10}{32} = \frac{5}{16}\).

Understanding Question 2

This involves finding the probability of throwing four ones in a row with a die, which is also a binomial problem with \(n = 4\) trials, \(p = \frac{1}{6}\) for each trial, and \(x = 4\) successes (ones).

Calculating Probability for Question 2

Using the binomial formula: \[ P(X = x) = \binom{n}{x}p^x(1-p)^{n-x} \] For 4 ones, \(\binom{4}{4} = 1\), so \(P(X = 4) = 1 \left(\frac{1}{6}\right)^4 = \frac{1}{1296}\).

Understanding Question 3

This involves finding the probability of experiencing at least one ski break in five competitions, where \(n = 5\), \(p = \frac{1}{120}\), and we need to find at least \(x = 1\) success. It is easier to compute the probability of \(x = 0\) successes and subtract it from one to find the probability of at least one ski break.

Calculating Probability for Question 3

Compute the probability of 0 ski breaks: \[ P(X = 0) = \binom{5}{0}\left(\frac{1}{120}\right)^0 \left(1 - \frac{1}{120}\right)^5 = 1 \cdot 1 \cdot \left(\frac{119}{120}\right)^5 = \left(\frac{119}{120}\right)^5 \approx 0.959\] Therefore, the probability of at least one ski break is \(b = 1 - 0.959 = 0.041\).

Key concepts

Probability Theory

Probability theory is a fundamental part of mathematics. It helps us understand how likely different outcomes are in any given situation. At its core, probability theory is about predicting chances, or how likely something is to happen.

We base probability on a sample space, which is a set of all possible outcomes. Important terms related to probability include "event," which is any subset of the sample space, and "outcome," which is a single result from the set of events.

Probability is often represented as a fraction between 0 and 1. A probability of 0 means the event will not happen, while a probability of 1 means the event is certain to happen.

There are two main rules in probability theory:

• The sum of all the probabilities of all possible outcomes must equal 1.
• The probability of an event happening is calculated by dividing the number of successful outcomes by the total number of possible outcomes.

Coin Flipping

Coin flipping is a classic example of a random experiment in probability theory. In a simple coin toss, there are two possible outcomes: heads or tails. Each outcome has an equal probability of occurring. This probability is \(\frac{1}{2}\) for either heads or tails.

When we flip a coin multiple times, we usually want to know the probability of getting a certain number of heads. This is where the binomial distribution comes into play. A binomial distribution models the probability of getting \(x\) successes (e.g., heads) in \(n\) trials, with \(p\) being the probability of success on a single trial.

For example, if we want to calculate the probability of getting exactly 2 heads in 5 coin flips, we would use the binomial formula:

• \[P(X = x) = \binom{n}{x}p^x(1-p)^{n-x}\]

In this case, \(n = 5\), \(x = 2\), and \(p = \frac{1}{2}\).

Dice Probability

Dice games are another excellent example of probability theory in action. When you roll a standard six-sided die, the probability of any specific number coming up is \(\frac{1}{6}\). This is because there are six possible outcomes, and each is equally likely.

The concept of dice probability is useful in many board games and gambling scenarios. Binomial distribution can be applied to find probabilities of multiple specific outcomes, like rolling a certain number multiple times. For example, if you roll a die four times, and want the probability of getting a "1" each time, this is analogous to four independent events, each with a success probability of \(\frac{1}{6}\).

Just as with coin flips, we can use the binomial formula to calculate these probabilities, setting \(n = 4\), \(x = 4\), and \(p = \frac{1}{6}\).

Events and Outcomes

In probability theory, clearly understanding "events" and "outcomes" is crucial. An event is a set of outcomes of an experiment or situation. An outcome is the result of a single trial of the experiment.

Events can be simple or compound:

• Simple events have only one outcome, like rolling a die and getting a "2".
• Compound events consist of two or more simple events, such as rolling a die twice and getting certain combinations, like "1" and then "3".

When analyzing probabilities, we look at whether events are independent or dependent. Independent events have no impact on each other, like two separate coin flips. Dependent events affect each other's outcomes, such as drawing cards without replacement from a deck.

The probability of an event is found by counting the successful outcomes and dividing by the total number of possible outcomes. This foundational understanding assists in solving more complex problems using distributions like the binomial.